Find the derivative of the following functions from first principle.
Question:

Find the derivative of the following functions from first principle.

(i) $x^{3}-27$

(ii) $(x-1)(x-2)$

(iii) $\frac{1}{x^{2}}$

 

(iv) $\frac{x+1}{x-1}$

Solution:

(i) Let $f(x)=x^{3}-27$. Accordingly, from the first principle,

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \frac{\left[(x+h)^{3}-27\right]-\left(x^{3}-27\right)}{h}$

$=\lim _{h \rightarrow 0} \frac{x^{3}+h^{3}+3 x^{2} h+3 x h^{2}-x^{3}}{h}$

$=\lim _{h \rightarrow 0} \frac{h^{3}+3 x^{2} h+3 x h^{2}}{h}$

$=\lim _{h \rightarrow 0}\left(h^{2}+3 x^{2}+3 x h\right)$

$=0+3 x^{2}+0=3 x^{2}$

(ii) Let $f(x)=(x-1)(x-2)$. Accordingly, from the first principle,

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \frac{(x+h-1)(x+h-2)-(x-1)(x-2)}{h}$

$=\lim _{h \rightarrow 0} \frac{\left(x^{2}+h x-2 x+h x+h^{2}-2 h-x-h+2\right)-\left(x^{2}-2 x-x+2\right)}{h}$

$=\lim _{h \rightarrow 0} \frac{\left(h x+h x+h^{2}-2 h-h\right)}{h}$

$=\lim _{h \rightarrow 0} \frac{2 h x+h^{2}-3 h}{h}$

$=\lim _{h \rightarrow 0}(2 x+h-3)$

$=(2 x+0-3)$

$=2 x-3$

(iii) Let $f(x)=\frac{1}{x^{2}}$. Accordingly, from the first principle,

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \frac{\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}}{h}$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{x^{2}-(x+h)^{2}}{x^{2}(x+h)^{2}}\right]$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{x^{2}-x^{2}-h^{2}-2 h x}{x^{2}(x+h)^{2}}\right]$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{-h^{2}-2 h x}{x^{2}(x+h)^{2}}\right]$

$=\lim _{h \rightarrow 0}\left[\frac{-h-2 x}{x^{2}(x+h)^{2}}\right]$

$=\frac{0-2 x}{x^{2}(x+0)^{2}}=\frac{-2}{x^{3}}$

(iv) Let $f(x)=\frac{x+1}{x-1}$. Accordingly, from the first principle,

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \frac{\left(\frac{x+h+1}{x+h-1}-\frac{x+1}{x-1}\right)}{h}$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{(x-1)(x+h+1)-(x+1)(x+h-1)}{(x-1)(x+h-1)}\right]$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{\left(x^{2}+h x+x-x-h-1\right)-\left(x^{2}+h x-x+x+h-1\right)}{(x-1)(x+h-1)}\right]$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{-2 h}{(x-1)(x+h-1)}\right]$

$=\lim _{h \rightarrow 0}\left[\frac{-2}{(x-1)(x+h-1)}\right]$

$=\frac{-2}{(x-1)(x-1)}=\frac{-2}{(x-1)^{2}}$

 

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